Electrostatics Question 368
Question: Two positive charges of 20 $ coulomb $ and $ Q\ coulomb $ are situated at a distance of $ 60cm $ . The neutral point between them is at a distance of $ 20cm $ from the $ 20coulomb $ charge. Charge $ Q $ is
Options:
A) $ 30C $
B) $ 40C $
C) $ 60C $
D) $ 80C $
Show Answer
Answer:
Correct Answer: D
Solution:
At neutral point $ k\times \frac{20}{{{(20\times {{10}^{-2}})}^{2}}}=k\times \frac{Q}{{{(40\times {{10}^{-2}})}^{2}}} $
therefore Q = 80 C
BETA
